1. CHEN 4460 – Process Synthesis, Simulation and Optimization
Dr. Mario Richard Eden Department of Chemical Engineering Auburn University
Lecture No. 8 – Mathematical Optimization
October 16, 2012
Contains Material Developed by Dr. Daniel R. Lewin, Technion, Israel

2. Lecture 10 – Objectives On completion of this part, you should:
Understand the different types of optimization problems and their formulation
Be able to formulate and solve a variety of optimization problems in LINGO

3. Optimization Basics What is Optimization? Examples
The purpose of optimization is to maximize (or minimize) the value of a function (called objective function) subject to a number of restrictions (called constraints).
Examples
1. Maximize reactor conversion
Subject to reactor modeling equations
kinetic equations
limitations on T, P and x

4. Optimization Basics Examples (Continued) 2. Minimize cost of plant
Subject to mass & energy balance equations
equipment modeling equations
environmental, technical and logical constraints

5. Optimization Basics Examples (Continued)
3. Maximize your grade in this course
Subject to extracurricular activities
full-time-job requirements
constant demand by other courses and/or your advisor/boss

6. Optimization Basics Formulation of Optimization Problems
min (or max) f(x1,x2,……,xN)
subject to g1(x1,x2,……,xN)≤0
g2(x1,x2,……,xN)≤0
gm(x1,x2,……,xN)≤0
h1(x1,x2,……,xN)=0
h2(x1,x2,……,xN)=0
hE(x1,x2,……,xN)=0
Feasibility
Any vector (or point) which satisfies all the constraints of the optimization program is called a feasible vector (or a feasible point)
The set of all feasible points is called feasibility region or feasibility domain
Any optimal solution must lie within the feasibility region!
Inequality Constraints
Equality Constraints

7. Optimization Basics Classification of Optimization Problems
Linear Programs (LP’s)
A mathematical program is linear if
f(x1,x2,……,xN) and gi(x1,x2,……,xN)≤0 are linear in each of their arguments:
f(x1,x2,……,xN) = c1x1 + c2x2 + …. cNxN
gi(x1,x2,……,xN) = ai1x1 + ai2x2 + …. aiNxN
where ci and aij are known constants.
Linear Programs (LP’s) can be solved to yield a global optimum. Solver routines can guarantee a truly optimal solution.

8. Optimization Basics Classification of Optimization Problems
Non-Linear Programs (NLP’s)
A mathematical program is non-linear if any of the arguments are non-linear. For example:
min 3x + 6y2
s.t. 5x + xy ≥ 0
Integer Programming
Optimization programs in which ALL the variables must assume integer values. The most commonly used integer variables are the zero/one binary integer variables.
Integer variables are often used as decision variables, e.g. to choose between two reactor types.
Non-Linear Programs (NLP’s) can be solved to yield a local optimum. Solver routines can not always guarantee a globally optimal solution.

9. Optimization Basics Classification of Optimization Problems
Mixed Integer Linear Programs (MILP’s)
Linear programs in which SOME of the variables are real and other variables are integers
Can be solved as individual LP’s by fixing the integer variables, thus a global optimum can be identified.
Mixed Integer Non-Linear Programs (MINLP’s)
Non-linear programs in which SOME of the variables are real and other variables are integers
Can be solved as individual NLP’s by fixing the integer variables, but depending on the nature of the NLP’s it may not be possible to find a global optimum.

10. Optimization Basics Formulation of Optimization Problems Step 1 Step 2
Determine the quantity to be optimized and express it as a mathematical function (this is your objective function)
Doing so also serves to define variables to be optimized (input variables or optimization variables)
Step 2
Identify all stipulated requirements, restrictions, and limitations, and express them mathematically. These requirements constitute the constraints
Step 3
Express any hidden conditions. Such conditions are not stipulated explicitly in the problem, but are apparent from the physical situation, e.g. non-negativity constraints

11. Optimization Example Hydrogen Sulfide Scrubbing
Two variable grades of MEA.
First grade consists of 80 weight% MEA and 20% weight water. Its cost is 80 cent/kg.
Second grade consists of 68 weight% MEA and 32 weight% water. Its cost is 60 cent/kg.
It is desired to mix the two grades so as to obtain an MEA solution that contains no more than 25 weight% water.
What is the optimal mixing ratio of the two grades which will minimize the cost of MEA solution (per kg)?

12. Variables (Basis 1 kg solution)
Optimization Example
Hydrogen Sulfide Scrubbing (Cont’d)
Objective function min z = 80x1 + 60x2
Constraints
Water content limitation 0.20x x2 ≤ 0.25
Overall material balance x1 + x2 =1
Non-negativity x1 ≥ 0
x2 ≥ 0
Variables (Basis 1 kg solution)
x1 Amount of grade 1 (kg)
x2 Amount of grade 2 (kg)
z Cost of 1 kg solution (cents)

13. Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Feasibility region
The set of points (x1, x2) satisfying all the constraints, including the non-negativity conditions.
Constraint on water content x x2 ≤ 0.25

14. Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Feasibility region
Non-negativity constraints x1  0 , x2  0

15. Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Feasibility region
Mass balance constraint x1 + x2 = 1

16. Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Feasibility region
Any optimal solution
must lie within the
feasibility region!

17. Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
By plotting objective function curves for arbitrary values of z (here 70 and 75) we can evaluate the results:
Optimal Point
Intersection between
x1 + x2 = 1
and
0.20x x2 = 0.25
In addition
70 < zmin < 75

18. Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Solving the two equations simultaneously yields the optimum amounts of the two MEA solutions along with the minimum cost of the mixture

19. Optimization Software
LINGO
Available on computers in Ross 306
To start entering a new optimization problem type:
Model:
Enter the objective function by typing:
min = ……; or max = ……;
Then enter the constraints.
Each line must end by a semi-colon ;
The final statement in the problem should be “end”

20. Optimization Software
Resolving MEA Example in LINGO
LINGO Input
Model:
min = 80*x1 + 60*x2;
0.2*x *x2 < 0.25;
x1 + x2 = 1;
x1 > 0;
x2 > 0;
end
LINGO Output
Rows= Vars= No. integer vars= ( all are linear)
Nonzeros= Constraint nonz= 6( are +- 1) Density=0.667
Smallest and largest elements in absolute value=
No. < : 1 No. =: 1 No. > : 2, Obj=MIN, GUBs <= 2
Single cols= 0
Optimal solution found at step:
Objective value:
Variable Value Reduced Cost
X E+00
X E+00
Row Slack or Surplus Dual Price E E
E+00
E+00
Value of objective function:
Value of variable x1:
Value of variable x2:

21. More Optimization Examples
Lab Experiment
Determine the kinetics of a certain reaction by mixing two species, A and B. The cost of raw materials A and B are 2 and 3 $/kg, respectively.
Let x1 and x2 be the weights of A and B (kg) to be employed in the experiment
The operating cost of the experiment is given by:
OC = 4(x1)2 + 5(x2)2
The total cost of raw materials for the experiment should be exactly $6. Minimize the operating cost!

22. More Optimization Examples
Lab Experiment (Cont’d)
LINGO Input
Model:
min = 4*x1^2 + 5*x2^2;
2*x1 + 3*x2 = 6;
x1 > 0;
x2 > 0;
end
LINGO Output
Rows= Vars= No. integer vars=
Nonlinear rows= 1 Nonlinear vars= 2 Nonlinear constraints= 0
Nonzeros= Constraint nonz= Density=0.583
Optimal solution found at step:
Objective value:
Variable Value Reduced Cost
X E+00
X E+00
Row Slack or Surplus Dual Price E
E-07
E+00
Value of objective function:
Value of variable x1:
Value of variable x2:

23. More Optimization Examples
Coal Conversion Plant
What are the optimal production rates of gaseous and liquid fuels that maximize the net profit of the plant?
x1 ≤ 4
2x2 ≤ 12
3x1 + 2x2 ≤ 18

24. More Optimization Examples
Coal Conversion Plant (Cont’d)
Objective function max z = 3x1 + 5x2
Constraints
Pretreatment capacity 3x1 + 2x2 ≤ 18
Gasification capacity x1 ≤ 4
Liquefaction capacity 2x2 ≤ 12
Non-negativity x1 ≥ 0
x2 ≥ 0

25. More Optimization Examples Maximum profit Z = 36 for x1 = 2 and x2 = 6
Coal Conversion Plant (Cont’d)
Graphical solution
Maximum profit Z = 36 for x1 = 2 and x2 = 6

26. More Optimization Examples
Coal Conversion Plant (Cont’d)
LINGO Input
Model:
max = 3*x1 + 5*x2;
3*x1 + 2*x2 <= 18;
x1 <= 4;
2*x2 <= 12;
x1 > 0;
x2 > 0;
end
LINGO Output
Rows= Vars= No. integer vars= ( all are linear)
Nonzeros= Constraint nonz= 6( are +- 1) Density=0.611
Smallest and largest elements in absolute value=
No. < : 3 No. =: 0 No. > : 2, Obj=MAX, GUBs <= 2
Single cols= 0
Optimal solution found at step:
Objective value:
Variable Value Reduced Cost
X E+00
X E+00
Row Slack or Surplus Dual Price
E+00 E E
E+00
E+00
Value of objective function: 36
Value of variable x1: 2
Value of variable x2: 6

27. More Optimization Examples
Methanol Delivery
Supply methanol for three Methyl acetate plants located in towns A, B, and C
Daily methanol requirements for each plant:
MeAc Plant location Tons/day A B C
Methanol production plants
MeOH plant
Capacity

28. More Optimization Examples
Methanol Delivery (Cont’d)
Shipping cost (100 $/ton)
Schedule the methanol delivery system to minimize the transportation cost
MeOH Plant
MeAc Plant A
MeAc Plant B
MeAc Plant C 1 2 5 3 8 11 6 15 4 7 9

29. More Optimization Examples
Methanol Delivery (Cont’d)
We define the transportation loads (tons/day) going from each MeOH plant to each MeAc plant as follows:
Total transportation cost (Z)
MeOH Plant
MeAc Plant A
MeAc Plant B
MeAc Plant C 1
X1A
X1B
X1C 2
X2A
X2B
X2C 3
X3A
X3B
X3C 4
X4A
X4B
X4C
Z = 2X1A + X1B + 5X1C + 3X2A + 0X2B + 8X2C + 11X3A + 6X3B + 15X3C + 7X4A + X4B + 9X4C

30. More Optimization Examples
Methanol Delivery (Cont’d)
Objective function min Z = 2X1A + X1B + 5X1C X2A + 0X2B X2C + 11X3A X3B + 15X3C X4A + X4B + 9X4C
Constraints
Availability/supply X1A + X1B + X1C = 7
X2A + X2B + X2C = 5
X3A + X3B + X3C = 3
X4A + X4B + X4C = 2
Requirements/demand X1A + X2A + X3A + X4A = 6
X1B + X2B + X3B + X4B = 1
X1C + X2C + X3C + X4C = 10

31. More Optimization Examples
Methanol Delivery (Cont’d)
Constraints
Non-negativity X1A ≥ 0
X1B ≥ 0
X1C ≥ 0
X2A ≥ 0
X2B ≥ 0
X2C ≥ 0
X3A ≥ 0
X3B ≥ 0
X3C ≥ 0
X4A ≥ 0
X4B ≥ 0
X4C ≥ 0

32. Mixed Integer Programs
Use of 0-1 Binary Integer Variables
Commonly used to represent binary choices
Dichotomy modeling

33. Mixed Integer Programs
The Assignment Problem
Assignment of n people to do m jobs
Each job must be done by exactly one person
Each person can at most do one job
The cost of person j doing job i is Cij
The problem is to assign the people to the jobs so as to minimize the total cost of completing all the jobs.
We can assign integer variables to describe whether a certain person does a certain job or not

34. Mixed Integer Programs
The Assignment Problem (Cont’d)
The event of person j doing job i is designated Xij
The objective function can be written as:
Since exactly one person will do job i, and each person at most can do one job, we get:

35. Mixed Integer Programs
Plant Layout – An Assignment Problem
Four new reactors R1, R2, R3 and R4 are to be installed in a chemical plant
Four vacant spaces 1, 2, 3 and 4 are available
Cost of assigning reactor i to space j (in $1000) is
Assign reactors to spaces to minimize the total cost
Reactor
Space 1
Space 2
Space 3
Space 4 R1 15 11 13 R2 12 17 R3 14 10 R4 16

36. Mixed Integer Programs
Plant Layout (Cont’d)
Let Xij denote the existence (or absence) of reactor i in space j, i.e. if Xij =1 then reactor i exists in space j
Objective function min Z = X X X X X X X X X X X X34
+ 17X X X X44

37. Mixed Integer Programs
Plant Layout (Cont’d)
Constraints
Each space must be assigned to one and only one reactor
X11 + X12 + X13 + X14 = 1
X21 + X22 + X23 + X24 = 1
X31 + X32 + X33 + X34 = 1
X41 + X42 + X43 + X44 = 1
Each reactor must be assigned to one and only one space
X11 + X21 + X31 + X41 = 1
X12 + X22 + X32 + X42 = 1
X13 + X23 + X33 + X43 = 1
X14 + X24 + X34 + X44 = 1

38. Mixed Integer Programs
Plant Layout (Cont’d)
Solve using LINGO
Optimal assignment policy
Reactor R1 in space 2
Reactor R2 in space 1
Reactor R3 in space 4
Reactor R4 in space 3
Minimum cost
Cost = = $49,000

39. Lecture 10 – Summary On completion of this part, you should:
Understand the different types of optimization problems and their formulation
Be able to formulate and solve a variety of optimization problems in LINGO

40. Other Business Homework Next Lecture – October 23
SSLW: 24.1 plus problems posted on class webpage
Due Tuesday October 23
LINGO software is available on class webpage as zip-file
Next Lecture – October 23
Heat and Power Integration (SSLW p )
Review of Midterm Exam
Thursday October 18 during lab sessions
You will get your tests back to look at during solution review